I have the following transition matrix for a Markov chain.

$\displaystyle \L A\;=\;\begin{pmatrix}0 \,&\, \frac{1}{3} \,&\, 0 \,&\, \frac{1}{3} \,&\, \frac{1}{3} \,&\, 0 \,&\, 0 \,&\, 0 \,&\, 0 \\

\frac{1}{5} & 0 & \frac{1}{5} & \frac{1}{5} & \frac{1}{5} & \frac{1}{5} & 0 & 0 & 0 \\

0 & \frac{1}{3} & 0 & 0 & \frac{1}{3} & \frac{1}{3} & 0 & 0 & 0 \\

\frac{1}{5} & \frac{1}{5} & 0 & 0 & \frac{1}{5} & 0 & \frac{1}{5} & \frac{1}{5} & 0 \\

\frac{1}{8} & \frac{1}{8} & \frac{1}{8} & \frac{1}{8} & 0 & \frac{1}{8} & \frac{1}{8} & \frac{1}{8} & \frac{1}{8} \\

0 & \frac{1}{5} & \frac{1}{5} & 0 & \frac{1}{5} & 0 & 0 & \frac{1}{5} & \frac{1}{5} \\

0 & 0 & 0 & \frac{1}{3} & \frac{1}{3} & 0 & 0 & \frac{1}{3} & 0 \\

0 & 0 & 0 & \frac{1}{5} & \frac{1}{5} & \frac{1}{5} & \frac{1}{5} & 0 & \frac{1}{5} \\

0 & 0 & 0 & 0 & \frac{1}{3} & \frac{1}{3} & 0 & \frac{1}{3} & 0

\end{pmatrix} $[/size]

This represents the problem:

A child is standing in a square on the grid (namely x).

Once time per second, he jumps to another square that is adjacent to x.

The grid looks like this:[/size]

Code:

` * - * - * - *`

| 1 | 2 | 3 |

* - * - * - *

| 4 | 5 | 6 |

* - * - * - *

| 7 | 8 | 9 |

* - * - * - *

[size=14]

My question is, how do I find the steady state vector for the Markov chain?